If $a = - 2$, the largest number in the set $\{ - 3a, 4a, \frac {24}{a}, a^2, 1\}$ is
$\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1$
Since all the numbers are small, we can just evaluate the set to be \[\{ (-3)(-2), 4(-2), \frac{24}{-2}, (-2)^2, 1 \}= \{ 6, -8, -12, 4, 1 \}\]
The largest number is $6$, which corresponds to $-3a$.
$\boxed{\text{A}}$