The equivalent resistance would decrease because you would have twice as many resistors in parallel when the switch is closed.

The original resistance is:

R + 1 / (1/(90+10) + 1/(10+90)) = R + 50Ω

after closing the swtich the resistance becomes:

R + [1/ ((1/90) + (1/10))] + [1/ ((1/10) + (1/90))] = R + 18Ω

Now the question says the equivalent resistance drops by 48% so multiply the equation for the original resistance by 0.48 and set it equal to the second equation:

0.48*(R + 50Ω) = R + 18Ω

now simplify and solve for R:

0.48R + 24Ω = R + 18Ω

R - 0.48R = 24Ω - 18Ω

0.52R = 6Ω

R = 11.54Ω

and you get that R = 11.54Ω