$$ T \propto R^{\frac{3}{2}} $$

For a satellite in a circular orbit with radius $$ R $$, the time period is given by

$$ T = 2\pi \sqrt{\frac{R^3}{GM}}. $$

If the radius is increased to $$ 1.03R $$, the new time period $$ T' $$ becomes

$$ T' = 2\pi \sqrt{\frac{(1.03R)^3}{GM}} = T \times (1.03)^{\frac{3}{2}}. $$

For a small change, we can approximate

$$ (1.03)^{\frac{3}{2}} \approx 1 + \frac{3}{2} \times 0.03 = 1 + 0.045 = 1.045. $$

This means the new time period is approximately 4.5% larger than the original time period.

Thus, the time period of the second satellite is larger by approximately 4.5%.