Assume there is no spring support at the left end
$\delta = \dfrac{w_oL^4}{8EI} = \dfrac{200(12^4)(12^3)}{8(1.5 \times 10^6)(144)}$
$\delta = 4.1472 \, \text{ in}$
Considering the spring reaction
$\delta - \delta_{spring} = \delta_R$
$4.1472 - \dfrac{R}{k} = \dfrac{RL^3}{3EI}$
$4.1472 - \dfrac{R}{300} = \dfrac{R(12^3)(12^3)}{3(1.5 \times 10^6)(144)}$
$4.1472 = 0.0079413R$
$R = 522.23 \, \text{ lb}$
$\delta_{spring} = \dfrac{R}{k} = \dfrac{522.23}{300}$
$\delta_{spring} = 1.74 \, \text{ in} \,$ answer