Answer
$m_2 = 36.1~kg$
Work Step by Step
We can find the acceleration of the system.
$y = \frac{1}{2}at^2$
$a = \frac{2y}{t^2} = \frac{(2)(12.0~m)}{(3.00~s)^2}$
$a = 2.67~m/s^2$
We can set up a force equation for each block. First we can set up a force equation for $m_1$
$\sum F = m_1~a$
$T - m_1~g~sin(\alpha) - m_1~g~cos(\alpha)~\mu_k = m_1~a$
$T = m_1~g~sin(\alpha) + m_1~g~cos(\alpha)~\mu_k + m_1~a$
$T = (20.0~kg)(9.80~m/s^2)~sin(53.1^{\circ}) + (20.0~kg)(9.80~m/s^2)~cos(53.1^{\circ})(0.40) + (20.0~kg)(2.67~m/s^2)$
$T = 257.2~N$
We can set up a force equation for $m_2$
$\sum F = m_2~a$
$m_2~g - T = m_2~a$
$m_2~(g-a) = T$
$m_2 = \frac{T}{g-a}$
$m_2 = \frac{257.2~N}{(9.80~m/s^2)-(2.67~m/s^2)}$
$m_2 = 36.1~kg$
