(c) Net force on $B\ F_{net} = \sqrt{F_{A}^{2} + F_{C}^{2}}$
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$F_{A} = \frac{15\ \times \ 12}{(3)^{2}} = 20\ dyne,\ F_{C} = \frac{12\ \times \ 20\ }{(4)^{2}} = 15dyne$
$\Rightarrow F_{net} = \sqrt{F_{A}^{2} + F_{C}^{2}} = \sqrt{(20)^{2} \times (15)^{2}} = 25dyne$