Finding a CPU's Capacitive Load - Computer Architecture
We can't calculate the actual value of the capacitive load without knowing the voltage and power but we don't need to, knowing the ratios of old voltage to new voltage, old capacitive load to new capacitive load and old power to new power is enough.

So we have your equation for the power consumption of a CMOS chip (assuming that most load is dynamic)

$$P = \frac{1}{2}cv^2f $$

Lets write that out for both the old and new chips.

$$P_\mathrm{old} = \frac{1}{2}c_\mathrm{old}v_\mathrm{old}^2f_\mathrm{old} $$

$$P_\mathrm{new} = \frac{1}{2}c_\mathrm{new}v_\mathrm{new}^2f_\mathrm{new} $$

Now lets write out what we have been told.

$$f_\mathrm{old} = 2\times10^9$$

$$p_\mathrm{new} = 0.45p_\mathrm{old}$$

$$c_\mathrm{new} = 0.6c_\mathrm{old}$$

$$v_\mathrm{new} = 0.65v_\mathrm{old}$$

Sustitute

$$P_\mathrm{old} = \frac{1}{2}c_\mathrm{old}v_\mathrm{old}^2 2\times 10^9$$

$$0.45P_\mathrm{old} = \frac{1}{2}0.6c_\mathrm{old}(0.65v_\mathrm{old})^2f_\mathrm{new} $$

$$0.45\frac{1}{2}c_\mathrm{old}v_\mathrm{old}^2 2\times 10^9 = \frac{1}{2}0.6c_\mathrm{old}(0.65v_\mathrm{old})^2f_\mathrm{new} $$

Cancel

$$0.45 \times 2\times 10^9 = 0.6(0.65)^2f_\mathrm{new} $$

Calculate numbers

$$ f_\mathrm{new} \approx 3.55 \times 10^9$$