Answer
The external-internal pressure difference is $6.0\times 10^4~Pa$.
Work Step by Step
The external-internal pressure difference is equal to the gauge pressure at a depth of 6.1 meters. We can find the gauge pressure at this depth.
$P_G = \rho~g~h$
$P_G = (1000~kg/m^3)(9.80~m/s^2)(6.1~m)$
$P_G = 6.0\times 10^4~Pa$
The external-internal pressure difference is $6.0\times 10^4~Pa$.
