I don't know the actual approach. I did it this way:
$2\cdot210=420$ (base 6)
$2\cdot103=210$ (base 6)
$3\cdot21=103\;$ (base 6)
Now $21$ (base 6) $= 13$ (base 10) = prime
So, the total number of prime factors is $3\ldots( 2, 3, 21)$
But, the problem I faced here is how the hell do you know the divisibility of the number at each stage? In base 10 it comes naturally. What to do?
On
The existence and unicity of any integer's factorization into prime factors is a property of $\Bbb N$ and has nothing to do with the representation into any given base.
I think you are fooled by your feeling that finding prime factors is easier in base 10, but consider this integer: 47955327990754321. Is it any easier to decide whether it is prime in base 10 than in say base 17 ?
You can apply some of divisibility tests used in base $10$ like for example you have that number is divisibile by $9$ if sum of digit is,same applies for $5$ in base $6$ or for any base $n$ you have that rule for $n-1$ also since $6=2\cdot 3$ you have that number is divisible by $2$ or $3$ if it's last digit is divisible by $2$ or $3$ ($2$ and $5$ in base $10$) also for $2^n$ and $3^n$ you have to check the last $n$ digits of number and check if that number from $n$ digits is divisible by $2^n$ or $3^n$.
You also have in base $10$ that number is divisible by $11$ if the sum of odd digits minus the sum of even digits is divisible by $11$ than so is the number,in base $6$ it would be for number $7$,same follows for any base $n$ and number $n+1$.Actually representing $7$ in base $6$ it's $11$.
Now the last digit of $420$ is 0 which is divisible by $2$ so $$420_6=(4\cdot 6^2+2\cdot 6)_{10}=(2\cdot(2\cdot 6^2+6))_{10}=210_6$$ If you had for example $320$ than it would be $$320_6=(3\cdot 6^2+2\cdot 6)_{10}=(2\cdot6^2+6^2+2\cdot 6)_{10}=(2\cdot(6^2+3\cdot 6+6))_{10}=(2\cdot(6^2+4\cdot 6))_{10}=(2\cdot 140)_6$$ This is almost the same like in base $10$ you have that $320=2\cdot 160$ where you know that $(6\cdot 2)_{10}=12_{10}$ also you know that $(4\cdot 2)_6=12_6$.It's a bit hard to get used to this but working few examples will help you.For example decomposing $124553_6$ as first last digit is divisible by 3 and since the first digit is smaller than $3$ we have that the number divided by $3$ is a $5$-digit number,and we have that first digit of the $5$-digit number must be greater than $1$ since otherwise that number multiplied by $3$ is $5$-digit and not $6$-digit number.If the digit is greater than $3$ than we have that that number multiplied by $3$ is bigger than $130000_6$ and our is smaller so we have that first digit is $2$,now the number $(3\cdot 20000)_6=100000_6$ so we have to find $\frac{124553_6-100000_6}{3}=\frac{24553_6}{3}$ again by the same process since the first digit is smaller than $3$ we have that the number is $4$-digit,if the first digit is smaller than $5$ than the number multiplied by $3$ is smaller than $23000$,so the first digit is $5$ now the number $(3\cdot 5000)_6=23000_6$ so now we have to find $\frac{124553_6-123000}{3}=\frac{1553_6}{3}$ now doing the same again we have that first digit is $2$ and $(3\cdot 200)_6=1000_6$ so we have to find $\frac{124553_6-124000}{3}=\frac{553_6}{3}$ now since first digit is bigger than $3$ we have that the number is $3$-digit and again by same process we find the first digit $1$ and we have that $(3\cdot 100)_6=300_6$ now we have to find $\frac{253_6}{3}$ by the same process we find the first digit is $5$ and we have that $(3\cdot 50)_6=230_6$ now we have to find $\frac{23_6}{3}=5_6$ now summing this all up we have that the number $124553_6=3\cdot(20000+5000+200+100+50+5)=3\cdot 25355$ Now since this number has sum of digits divisible by $5$ you can divide it by $5$(almost identical process just instead of $3$ you use $5$ $25355_6/5=3311$ and now since the sum of odd minus sum of even is 0 you have that the number is divisible by $11_6=7_{10}$ now this is easy $3311=330\cdot 11+11=(11\cdot331)_6$ and $331$ is prime.
Also the number of factors must be the same in base $10$ and base $6$ since if you have a prime factor $x$ in base $10$ than you can write the same factor in base $6$.
This post seems a bit vague,sorry about that