$\sqrt{(121)_{r}}=11_{r}$
$\sqrt{(1\times r^{0})+(2\times r^{^{1}})+(1\times r^{2})}=(1\times r^{0})+(1\times r^{1})$
$\sqrt{(1+r)^{^{2}}}=1+r$
$1+r=1+r$
So any integer $r$ satisfies this but $r$ must be greater than $2$ as we have $2$ in $121$ and radix must be greater than any of the digits. (D) is the most appropriate answer