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where X1,..Xn are i.i.d with $ N(0, {\sigma}^2) $. Let $\hat{\sigma_{n}}$ = 1/n $ \Sigma_{i=1}^{n} {(X_{i})^2} $.

In the hints the law of large numbers has been used for $Y_i= (X_i)^2$ and $E[(X_i)^2]$ and $E[(X_i)^4]$ has been calculated and I do not understand how this related to using the law of large numbers. Further I am unsure as to how $E[(X_i)^2]$ has been calculated to equal ${\sigma}^2 $ ?

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    It is not unless $\mu = 0$ – 2017-02-25
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    @V.V Yes that's true I have edited it I made a typo there. Would you be able to help with my question ? – 2017-02-25

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Recall that if $X_1, X_2,...$ are i.i.d with $E|X|<\infty$ then by the WLLN $$ \frac{1}{n}\sum_{i=1}^nX_i \xrightarrow{p}EX. $$ In your case denote $Y_i = X_i^2$, then $Y_1, Y_2.,,$ are i.i.d with $EY_i=EX_i^2=Var(X_i)=\sigma^2$, hence by WLLN

$$ \frac{1}{n}\sum_{i=1}^nX^2_i = \frac{1}{n}\sum_{i=1}^nY_i \xrightarrow{p}EY =\sigma^2\, . $$ The finite variance ("second moment") of $Y_i$ is not strictly necessary for the WLLN to hold, but usually it makes the proofs easier and the convergence faster.