Let $X$ be uniformly distributed on the ball $B_{\sqrt n}$ of radius $\sqrt n$ in $\mathbb R^n$. Then $$X=RY,$$ where $R:=|X|/\sqrt n$ and $Y:=\sqrt n\,X/|X|$ is uniformly distributed on the sphere $S_{\sqrt n}$ of radius $\sqrt n$ in $\mathbb R^n$.

Note that $0\le R\le1$ and hence $E(X\cdot t)^2=ER^2(Y\cdot t)^2\le E(Y\cdot t)^2$ for all $t\in\mathbb R^n$. So, $\|X\|_{\psi_2}\le\|Y\|_{\psi_2}$.

Also, you know that $\|Y\|_{\psi_2}\le C$. Thus, $$\|X\|_{\psi_2}\le C,$$ as desired.

Let $X$ be uniformly distributed on the ball $B_{\sqrt n}$ of radius $\sqrt n$ in $\mathbb R^n$. Then $$X=RY,$$ where $R:=|X|/\sqrt n$ and $Y:=\sqrt n\,X/|X|$ is uniformly distributed on the sphere $S_{\sqrt n}$ of radius $\sqrt n$ in $\mathbb R^n$.

Note that $0\le R\le1$ and hence $E\exp\{c(X\cdot t)^2\}=E\exp\{cR^2(Y\cdot t)^2\}\le E\exp\{c(Y\cdot t)^2\}$ for all real $c>0$ and $t\in\mathbb R^n$. So, $\|X\|_{\psi_2}\le\|Y\|_{\psi_2}$.

Also, you know that $\|Y\|_{\psi_2}\le C$. Thus, $$\|X\|_{\psi_2}\le C,$$ as desired.