Whether b equals 2 or 4 depends on your representation of input size.

If the size of the input is n (i.e., the size of one dimension of each input array), then the running time of the naive algorithm is O(n³) and each square array in each subproblem has dimension n/2. Thus b is 2 and the running time for Strassen's is O(n^log₂7).

If the size of the input is the number of elements in each array m=n², then the running time of the naive algorithm is O(m m^1/2) = O(m^3/2) and each square array in each subproblem has m/4 elements in it. Thus b is 4 and the running time for Strassen's is O(m^log₄7).

Relative to their representations of input size, these running times are identical since you can map between them by changing the representation with m=n²:

For the naive algorithm, O(n³)=O((m^1/2)³) = O(m^3/2).

For Strassen's, O(n^log₂7) = O((m^1/2)^log₂7) = O(m^{(log_2 7)/2})=O(m^{(log₂7)/(log₂4)}) = O(m^log₄7).