Yes. For each $x\in R$ let $a(x)$ and $b(x)$ be the number of $y$ such that $xy=0$ and the number of $y$ such that $yx=0$, respectively. These are also the orders of the groups $R/xR$ and $R/Rx$.
For every $x$ the left ideal $Rx$ is a right ideal containing $x$ and therefore contains $xR$, whence $b(x)\le a(x)$. On the other hand the sum of $a(x)$ over all $x$ is clearly the same as the sum of $b(x)$ (both are the number of pairs $(x,y)$ such that $xy=0$). Therefore $a(x)=b(x)$ and $Rx=xR$. This shows that $xR$ is a left ideal for every $x$, and it follows that every right ideal is a left ideal.