Arate is equal to 0.0200 m/s. if [a] = 0.100 m and rate = k[a]2[b]2, what is the new rate if the concentration of [a] is increased to 0.200 m?

The rate equation is r = k[a]²[b]².

Take logarithms of both sides.
ln r = ln k + 2 ln [a] + 2 ln [b]

When [a] = 0.0.100 m, r₁ = 0.0200 m/s.
Therefore
ln(0.0200) = ln k + 2 ln (0.0100) + 2 ln [b]          (1)

When a = 0.200 m, obtain the new rate, r₂, as
ln r₂ = ln k + 2 ln (0.0200) + 2 ln [b]                 (2)

Subtract (1) from (2) to obtain
ln r₂ - ln(0.0200) = 2{ln(0.0200 - ln(0.0100)} 
ln(r₂/0.0200) = 2 ln(0.0200/0.0100)
                       = 2 ln(2)
                       = ln(2²) = ln(4)
Therefore
r₂/0.0200 = 4
r₂ = 0.0800  m/s

0.0800 m/s