Based on circumferential stress (tangential):

$\Sigma F_V = 0$

$F = 2T$

$p(DL) = 2(\sigma_t \, L_t)$

$\sigma_t = \dfrac{pD}{2t}$

$60 = \dfrac{p(450)}{2(20)}$

$p = 5.33 \, \text{MPa}$

Based on longitudinal stress:

$\Sigma F_H = 0$

$F = P$

$p (\frac{1}{4} \pi D^2) = \sigma_l(\pi D)$

$\sigma_l = \dfrac{pD}{4t}$

$140 = \dfrac{p(450)}{4(20)}$

$p = 24.89 \, \text{MPa}$

Use $p = 5.33 \, \text{MPa}$ *answer*