Force F acting on the shaded region

$F = ( \, f_b \, / \, c \, ) \, A' \, \bar{y} \, '$

where

$f_b = 1200 \, \text{ psi}$
$c = 4 \, \text{ in.}$

$A' = 6(2) = 12 \, \text{ in.}^2$

$\bar{y} \, ' = 3 \, \text{ in.}$

Thus,

$F = (1200/4)(12)(3)$

$F = 10 800 \, \text{lb}$

$F = 10.8 \, \text{kips}$ *answer*

Moment of F about the neutral axis

$M = f_b \, I' \, / \, c$

where

$I' = I + Ad^2$
$I' = \dfrac{6(2^3)}{12} + (6 \times 2)(3^2)$

$I' = 112 \, \text{in}^4$

Thus,

$M = 1200(112)/4$

$M = 33\,600 \, \text{lb}\cdot\text{in}$

$M = 2800 \, \text{lb}\cdot\text{ft}$ *answer*