Load sharing: 600 and 500 MVA, 900 MVA load

Taking "increased to 100 MVA" as increased by 100 MVA (total 1000 MVA), and treating MVA as MW (unity power factor), both units share the increase in proportion to $P_r/R$.

Method: a unit with droop $R$ (pu on its own rating $P_r$) and rated frequency $f_0$ changes its output by

$$\Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f$$

where $K = P_r/(R f_0)$ is its stiffness in MW/Hz. All units run at the same frequency, so $\Delta P_L = -\Delta f \sum K$.

Data: Unit 1: 600 MVA, 4 %, supplying 500; Unit 2: 500 MVA, 6 %, supplying 400; $f_0 = 60$ Hz; $\Delta P_L = 100$.

$$\begin{aligned} K_1 &= \frac{600}{0.04 \times 60} = 250\ \text{MVA/Hz} \\ K_2 &= \frac{500}{0.06 \times 60} = 138.89\ \text{MVA/Hz} \\ \Delta f &= -\frac{100}{388.89} = -0.2571\ \text{Hz} \\ f_{new} &= 60 - 0.2571 = 59.74\ \text{Hz} \\ \Delta P_1 &= 250 \times 0.2571 = 64.29 \\ \Delta P_2 &= 138.89 \times 0.2571 = 35.71 \end{aligned}$$

Answer: New frequency ≈ 59.74 Hz; Unit 1 generates 564.29 MVA (MW) and Unit 2 435.71 MVA (MW).