How can I roughly estimate the load in this case?
3 &times; 1.5 V, 900 mAh gives 4.5 V, 900 mAh.

Run time = 4 hours so \$ I = \frac {900\ \text {mAh}}{4 \ \text h} = 225 \ \text {mA} \$.

The effective load resistance is given by \$ R = \frac {V}{I}  = \frac {4.5}{225} \$ which will be left as an exercise for the reader.