The entire question is in the title, however, I am posting this because I solved it and something seems off, hence the "solution-verification" tag this is posted under.
My solution:
We know that if X is a random variable following the Gamma distribution then $E(X)=\alpha \beta$, so in our case, we know that $E(X)=5\beta$, if we have a random variable X with $\alpha$ and $\beta$ as parameters then then random variable $X^2$ will have $\alpha^2$ and $\beta^2$ (here is where I think I might be thinking wrong as this basically would result in $E(X^2)=E^2(X)$ which is a common error, however, I came to this conclusion by thinking of any Gamma function and squaring it which would result in $\alpha$ and $\beta$ becoming squared), thus we have the following:
$25\beta^2=11.7187500$
$\beta=\sqrt{\frac{11.7187500}{25}}$
Any help in fixing my mistakes or confirming my solution would be appreciated.
If $X \sim \operatorname{Gamma}(\alpha,\beta)$ such that $\operatorname{E}[X] = \alpha \beta$, then $\operatorname{Var}[X] = \alpha \beta^2$. Consequently, $$\operatorname{E}[X^2] = \operatorname{Var}[X] + \operatorname{E}[X]^2 = \alpha \beta^2 + \alpha^2 \beta^2 = \alpha (\alpha+1) \beta^2,$$ not $\alpha^2 \beta^2$.
Alternatively, we can compute the moments directly:
$$\operatorname{E}[X^k] = \int_{x=0}^\infty x^k \frac{x^{\alpha-1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)} \, dx = \frac{\beta^k \Gamma(\alpha + k)}{\Gamma(\alpha)} \int_{x=0}^\infty \frac{x^{\alpha + k - 1} e^{-x/\beta}}{\beta^{\alpha + k} \Gamma(\alpha+k)} \, dx,$$ and here we note that the integrand has been carefully rewritten so that it corresponds to a gamma density with shape parameter $\alpha + k$ and scale parameter $\beta$; thus it integrates to $1$ over its support, and we have $$\operatorname{E}[X^k] = \frac{\beta^k \Gamma(\alpha + k)}{\Gamma(\alpha)}.$$ Then for $k = 1$, we recover the mean $$\operatorname{E}[X] = \beta \frac{\Gamma(\alpha+1)}{\Gamma(\alpha)} = \alpha \beta,$$ and for $k = 2$, we obtain the claimed formula above: $$\operatorname{E}[X^2] = \beta^2 \frac{\Gamma(\alpha + 2)}{\Gamma(\alpha)} = \alpha (\alpha + 1) \beta^2.$$