Given data:
Assume $I_{\text {TAL }}=0$

$$
\begin{aligned}
& R_{i d}=2 \mathrm{M} \Omega \\
& A_{d m}=-500
\end{aligned}
$$

$\mathrm{CMRR}=500$
Neglect $r_b, r_\mu$, and $r_o$
Under zero-signal condition, circuit is:
The differential input resistance of the emitter coupled pair circuit is:

$$
\begin{aligned}
& R_{i d}=2 r_{\mathrm{x}} \\
& 2 \mathrm{M}=2 r_{\mathrm{x}} \\
& r_{\mathrm{x}}=1 \mathrm{M} \Omega
\end{aligned}
$$

$$
\begin{aligned}
& r_s=\frac{\beta}{g_m} \\
& 1 \mathrm{M}=\frac{200}{g_m} \\
& g_m=\frac{200}{1 \times 10^6} \\
& g_m=2 \times 10^{-4} \\
& g_m=0.2 \mathrm{~mA} / \mathrm{V} \\
& I_C=I_{C 1}=I_{C 2}
\end{aligned}
$$


We know that $g_m=\frac{I_C}{V_T}$

$$
\begin{aligned}
& I_C=g_m V_T \\
& V_T=26 \mathrm{mV} \\
& I_C=0.2 \times 10^{-3} \times 26 \times 10^{-3} \\
& I_C=5.2 \times 10^{-6} \\
& I_C=5.2 \mu \mathrm{~A}
\end{aligned}
$$


The differential mode gain $A_{2 m}$ for the emitter coupled pair circuit is:

$$
\begin{aligned}
& A_{i m}=-g_m R_C \\
& -500=-0.2 \mathrm{~m} \times R_C
\end{aligned}
$$

$$
\begin{aligned}
& R_C=\frac{-500}{-0.2 \times 10^{-3}} \\
& R_C=2.5 \times 10^6 \\
& \therefore R_C=2.5 \mathrm{M} \Omega
\end{aligned}
$$


The DC voltage drop across $R_C$ is

$$
\begin{aligned}
& I_C R_C=\left(5.2 \times 10^{-6}\right)\left(2.5 \times 10^6\right) \\
& I_C R_C=13 \mathrm{~V}
\end{aligned}
$$


When the bases of transistors $Q_1$ and $Q_2$ are grounded then, we have

$$
V_{C B}=V_{C E}-V_{B E}
$$


For a transistor $V_{C E}$ of 1 V

$$
\begin{array}{ll}
V_{C B}=1-0.7 & \left(\because V_{B E}=0.7 \mathrm{~V}\right) \\
V_{C B}=0.3 \mathrm{~V} &
\end{array}
$$


Writing KVL in the input loop of transistor $Q_1$, we get:

$$
\begin{aligned}
& V_{C C}=V_{C B}+I_C R_C \\
& V_{C C}=0.3+13 \\
& \therefore V_{C C}=13.3 \mathrm{~V}
\end{aligned}
$$


The common mode gain $A_{c m}$ for the emitter coupled pair circuit is:
$$
A_{c m}=-\frac{g_m R_C}{1+2 g_m R_{\text {Tail }}}
$$


Since $2 g_m R_{\text {TALL }} \gg 1$, we can neglect 1 compared to $2 g_m R_{\text {TAIL }}$

$$
\begin{aligned}
& A_{c m} \approx-\frac{R_C}{2 R_{\text {Tail }}} \\
& \mathrm{CMRR}=\frac{A_{a m}}{A_{c m}} \\
& \mathrm{CMRR}=\frac{-g_m R_C}{\left(-\frac{R_C}{2 R_{\text {Tail }}}\right)} \\
& \mathrm{CMRR}=2 g_m R_{\text {Tail }} \\
& 500=2 \times 0.2 \times 10^{-3} \times R_{\text {TAII }} \\
& R_{\text {TALI }}=\frac{500}{4 \times 10^{-4}} \\
& R_{\text {TALI }}=125 \times 10^4 \\
& \therefore R_{\text {TALI }}=1.25 \mathrm{M} \Omega
\end{aligned}
$$