jmc

If $a\ge 4$, then $a^3+b^2+c>a^3\ge 4^3>50$. But we want $a^3+b^2+c\le 50$, so we must have $a=2$. Now we substitute $a=2$ into $a^3+b^2+c\le 50$, which gives $b^2+c\le 42$. Since $b^2<42$, we know that $b$ must be one of 2, 4, or 6.

When $b=2,$ $c\le 38$. There are 19 even positive integers less than or equal to 38, namely $2\times 1$, $2\times 2$, $\dots$, $2\times 19$.

When $b=4,$ $c\le 26$. There are 13 even positive integers less than or equal to 26.

When $b=6,$ $c\le 6$. There are 3 even positive integers less than or equal to 6.

Thus the answer is $19+13+3=\boxed{35}$.