$M_{max} = \frac{1}{4}PL = \frac{1}{4}P(12) = 3P \, \text{ lb}\cdot\text{ft}$

$V_{max} = \frac{1}{2}P$

$I = \frac{1}{12} (6)(12^3) = 864 \, \text{ in}^4$

From allowable flexural stress

$f_b = \dfrac{Mc}{I}$

$1200 = \dfrac{(3P \times 12)(6)}{864}$

$P = 4800 \, \text{ lb}$

Strength of bolt

$R = \dfrac{VQ_\text{1st plank}}{I}s$

$R = \dfrac{\frac{1}{2}(4800) \, [ \, 4(6)(4) \, ]}{864}(12)$

$R = 3200 \, \text{ lb}$

Normal force

$R = \mu N$

$3200 = 0.40N$

$N = 8000 \, \text{ lb}$

From tensile stress of bolt:

$\sigma = \dfrac{\text{Force}}{\text{Area}}$

$20\,000 = \dfrac{8000}{\frac{1}{4}\pi d^2}$

$d = 0.7136 \, \text{ in}$ *answer*