How to prove that $J(M_n(R))=M_n(J(R))$? How to prove that $J(M_n(R))=M_n(J(R))$?
Here $M_n(R)$ is the ring of matrices of size $n^2$ over the ring $R$. And $J(M_n(R))$ is a two-sided ideal of the ring $M_n(R)$.
Take any matrix $A$ in $J$. Multiply it on the left by matrix that has $1$ on $i$-th place on the diagonal and zeroes elsewhere. The result will be a matrix $A'$ that has $i$-th row of $A$ and zeroes elsewhere. Since $J$ was double sided ideal, it's also in $J$. Now multiply $A'$ on the right by the matrix that has $1$ on $j$-th place on the diagonal and zeroes elsewhere. You'll get matrix which has $j$-th column of the matrix $A'$, so it will precisely be a matrix that has the same element as $A$ on the $i, j$ position, and zeroes elsewhere. Since $J$ was doublesided ideal, it will also be in $J$.

Do you see how to continue?
Take any matrix $A$ in $J$. Multiply it on the left by matrix that has $1$ on $i$-th place on the diagonal and zeroes elsewhere. The result will be a matrix $A'$ that has $i$-th row of $A$ and zeroes elsewhere. Since $J$ was double sided ideal, it's also in $J$. Now multiply $A'$ on the right by the matrix that has $1$ on $j$-th place on the diagonal and zeroes elsewhere. You'll get matrix which has $j$-th column of the matrix $A'$, so it will precisely be a matrix that has the same element as $A$ on the $i, j$ position, and zeroes elsewhere. Since $J$ was doublesided ideal, it will also be in $J$.

Do you see how to continue?