Answer
$b = 4$ units
Work Step by Step
$c^{2} = a^{2} + b^{2} - 2(a)(b)cos(C)$
$(2\sqrt {3})^{2} = (2)^{2} + (b)^{2} - 2(2)(b)cos(60)$
$(4 \times 3) = 4 + b^{2} - 4bcos(60)$
$12 = 4 + b^{2} - 4bcos(60)$
$8 = b^{2} - 4bcos(60)$
$b^{2} - (4cos(60))b - 8 = 0$
Let $b = x$
$x^{2} - (4cos(60))x - 8 = 0$
$x^{2} - 2x - 8 = 0$
$(x-4)(x+2) = 0$
$x = 4, -2$
$b = 4, -2$
Since $b \ne -2$, then the only answer is $b = 4$ units.
