#### Answer

$a = 1.77~m/s^2$

#### Work Step by Step

Note that the 1.0-kg block has the same magnitude of acceleration as the 2.0-kg block. We can set up a force equation for the 1.0-kg block. Let $m_1$ be the mass of this block.
$\sum F = m_1~a$
$T-F_f = m_1~a$
$T-m_1~g~\mu_k = m_1~a$
$T = m_1~g~\mu_k + m_1~a$
We can use the expression for tension $T$ in the force equation for the 2.0-kg block. Let $m_2$ be the mass of this block.
$\sum F = m_2~a$
$F - T - F_{f1}- F_{f2} = m_2~a$
$F - (m_1~g~\mu_k + m_1~a) - m_1~g~\mu_k- (m_1+m_2)~g~\mu_k = m_2~a$
$F - 3m_1~g~\mu_k - m_2~g~\mu_k = (m_1+m_2)~a$
$a = \frac{F - 3m_1~g~\mu_k - m_2~g~\mu_k}{m_1+m_2}$
$a = \frac{20~N - (3)(1.0~kg)(9.80~m/s^2)(0.30) - (2.0~kg)(9.80~m/s^2)(0.30)}{1.0~kg+2.0~kg}$
$a = 1.77~m/s^2$