Answer
Satellite B has a greater total energy than satellite A by $~~1.14\times 10^8~J$
Work Step by Step
Let $R_E$ be the Earth's radius.
The total energy of an orbiting satellite at a height $h$ above the Earth's surface is $E = -\frac{GMm}{2(R_E+h)}$
We can see that as the value of $h$ increases, the total energy becomes less negative. (That is, the magnitude decreases as the value moves closer to zero.)
Since satellite B is at a higher altitude than satellite A, satellite B has a greater total energy than satellite A.
We can find the difference in the total energy of each satellite:
$E_B-E_A = -\frac{GMm}{2(R_E+h_B)}- [-\frac{GMm}{2(R_E+h_A)}]$
$E_B-E_A = \frac{GMm}{2(R_E+h_A)} -\frac{GMm}{2(R_E+h_B)}$
$E_B-E_A = \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)(14.6~kg)}{(2)(6.37\times 10^6~m+6.37\times 10^6~m)} - \frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)(14.6~kg)}{(2)(6.37\times 10^6~m+1.911\times 10^7~m)}$
$E_B-E_A = 2.28\times 10^8~J - 1.14\times 10^8~J$
$E_B-E_A = 1.14\times 10^8~J$
Satellite B has a greater total energy than satellite A by $~~1.14\times 10^8~J$
