Transformer circuit, equivalent impedance
To maximize power into the resistors the inductive reactance needs to be cancelled by the capacitive reactance. This is because the circuit is series-tuned (despite the presence of the transformer).

You know the frequency (1kHz) and you know the inductance (1uH) and the natural resonant frequency for an LC circuit (series or parallel) is: -

\$F_N = \dfrac{1}{2\pi\sqrt{LC}}\$

So then you transpose this to get the required capacitor value that would tune the circuit.

But your cap value is 1uF and this needs to be transformed to the value you have just worked out using the resonance formula. Does it need to increase from 1uF or decrease?

I suspect it needs to increase and so you have to decide by how much. Let's say it needs to increase ten thousand times to (say) 10,000\$\mu F\$, what should the turns ratio be to obtain this impedance transformation?

The hint in your question tells you that it is the square of the turns ratio that governs how this works so, if the required impedance transformation is 10,000 then the actual turns ratio is the square root of this.

I'll leave you to decide whether the turns ratio is 10,000:1 from left to right or 1:10,000. By the way 10,000 isn't the right answer - you need to calculate this.