Explanation
According to given data
We know that,
$$mvr=\dfrac{nh}{2\pi }$$
Or,
$$qvB=\dfrac{m{{v}^{2}}}{r}$$
So we find,
$$qB=\dfrac{mv}{r}$$
$$qB\left( \dfrac{nh}{2\pi mv} \right)=mv$$
$$ \left( \dfrac{1}{2}m{{v}^{2}} \right)=\dfrac{nhqB}{4\pi m} $$
$$ E=n\left( \dfrac{hqB}{4\pi m} \right) $$
This is the required solution.
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