The probability of getting an one is 1/6. It follows a binomial distribution with $n=1200$ and p=1/6.
So $\mu=np=1200*(1/6)=200$ and $\sigma=\sqrt{np(1-p)}=\sqrt{1200*(1/6)*(5/6)}=12.91$.
Standardizing the values then using the Z-table:
$z1=\frac{180-\mu}{\sigma}\rightarrow \frac{180-200}{12.91} \rightarrow -1.55$
$z2=\frac{220-\mu}{\sigma}\rightarrow \frac{220-200}{12.91} \rightarrow 1.55$
$P(-1.55 \leq Z \leq 1.55) = 0.9392$. So the probability of having $180 \leq X \leq 220 $ is 93.92%.
Is my answer correct?
This is correct through the calculation of the z-scores $z_1$ and $z_2$. When you look up the standard normal values that correspond to them, the tables or calculators provide $P(Z\leq z)$ which is usually shown as a shaded region under the bell curve from $z$ all the way to the left.
in this case, $P(Z\leq 1.55) = .9394$ and this represents a close approximation to $P(X\leq 220)$.
By the symmetry of the bell curve you can work out (or just look up) that
$P(Z\leq -1.55) = 1- P(Z\leq 1.55) = .0606$
Finally based on this, $P(180\leq X \leq 220) \approx P(-1.55\leq Z \leq 1.55) = .9394-.0606 = 87.9\%$