Write $n$ as $a_0 + 10a_1 + 100a_2 + \dots$, where $a_0, a_1, a_2, \dots$ are the digits of $n$. Then the stated equation is equivalent to
$$a_0 + 10a_1 + 100a_2 + \dots = a_0 + a_1 + a_2 + \dots + a_0^2$$
or
$$9a_1 + 99a_2 + 999a_3 + \dots = a_0^2.$$
The right hand side is at most $9^2 = 81 < 99$. Therefore, $a_2, a_3, \dots$ have to be 0 (otherwise, the left hand side would be strictly greater than the right hand side). Hence we obtain
$$9a_1 = a_0^2.$$
It follows that $a_0$ must be divisible by 3 (since $9a_1$ is), which leaves us with the possibilities $a_0 = 3$ ($a_1 = 1$), $a_0 = 6$ ($a_1 = 4$) and $a_0 = 9$ ($a_1 = 9$). So there are three solutions: 13, 46 and 99.