Rivet capacity in terms of shear (double shear)
$R_s = 2(A_s \, \tau) = 2 \, [ \, \frac{1}{4}\pi (22^2)(100) \, ]$
$R_s = 24\,200\pi \, \text{ N } = 24.2\pi \, \text{ kN } = 76.03 \text{ kN}$
Rivet capacity in terms of bearing (use σb = 280 MPa)
$R_b = A_b \, \sigma_b = [ \, 22(10) \, ] (280)$
$R_b = 61\,600 \, \text{ N } = 61.6 \, \text{ kN}$
Use R = 61.6 kN for safe value of R
Moment of area
$Q_A = 2(2430)(491.1) + 300(10)(515)$
$Q_A = 3,931,746 ~ \text{mm}^3$
Spacing of rivets
$s = \dfrac{RI}{VQ_A}$
$s = \dfrac{61.6(4770 \times 10^6)}{450(3,931,746)}$
$s = 166 ~ \text{mm}$ answer