Answer
See the detailed answer below.
Work Step by Step
We know that a resistor's voltage and current are in phase while the current leads the voltage of a capacitor by 90$^\circ$.
And we are given that the transformer voltage produces a 6.0 V rms output that leads the input voltage by 45$^\circ$.
This means that we need to make $V_R=V_C$ to make $V_R$ and $I$ leads the input voltage ($\varepsilon$) by $45^\circ$.
$$V_R=V_C\\
\color{red}{\bf\not} IR= \color{red}{\bf\not} IX_C\\
R=X_C$$
$$R=\dfrac{1}{2\pi f C}\tag 1$$
Thus, the phase angle is then
$$\phi=\tan^{-1}\left[ \dfrac{|X_L-X_C|}{R}\right]=\tan^{-1}\left[ \dfrac{X_C}{R}\right]=\bf 45^\circ$$
See the first graph below of the phasor diagram for an $RC$ circuit.
We designed the circuit as shown below in the second figure, so we need to make sure that the output voltage is 6 V.
$$(V_R)_{\rm rms}=I_{\rm rms}R=\dfrac{\varepsilon_{\rm rms}R}{\sqrt{X_C^2+R^2}}$$
where $X_C=R$, so
$$(V_R)_{\rm rms}=I_{\rm rms}R=\dfrac{\varepsilon_{\rm rms} }{\sqrt{2}}=\dfrac{12}{\sqrt2}=\bf 8.49\;\rm V\gt 6\;\rm V$$
which is greater than the needed output voltage of 6 V.
Now we can reduce the output voltage by using two in-series resistors that have a net resistance as the original resistor $R$, as shown in the third figure below; where $R_1=R_2=R$.
So, $V_1=IR_1$, and $V_2=IR_2$
So,
$$\dfrac{V_2}{V_R}=\dfrac{V_2}{V_1+V_2}=\dfrac{R_2}{R_1+R_2}=\dfrac{R_2}{R}$$
Hence,
$$R_2=\dfrac{RV_2}{V_R}$$
where $V_2$ is assumed to be exactly 6 V while $V_R=12/\sqrt2$,
$$R_2=\dfrac{ 6\sqrt2}{12}R$$
$$\boxed{R_2= 0.7071R}$$
And hence,
$$\boxed{R_1= 0.2929R}$$
We can see that any values that obey these formulas will give us the needed circuit.