Iref = 0.5mA
0.7 = 0.025 ln (1mA / Is)
to get the 0.5 mA reference current, we need a Vbe = 0.025 ln (0.5mA / Is)
Vbe = 0.025 ln ((0.5mA/ 1mA)  *  (1mA / Is) ) = 0.025 (ln (0.5/1) + ln(1mA/ Is) )
    = 0.025 (-0.693 + 0.7/0.025) = - 0.017 + 0.7  = 0.68v
Io = 0.5mA = (5 - 0.68) / R
so R = 8.6K

now, using this reference voltage Vbe, we put four of the transistors in parallel on the
output.  Because Beta is very large, the current gain is extremely close to 4.
Note that because the problems calls for the use of transistors for which Vbe=0.7v at 1mA,
you cannot simply change the size of the emitter-base junction to scale the output...

Rid = (1+B)*2*Re
re = 25mV / 50uA = 500 ohms
so
Rid = 101 * 1000 = 101Kohms
Ro = ro4 || ro2 = ro / 2
ro = VA / Ic = 160v / 50uA = 3.2 Meg
thus, Ro = 1.6Meg
Gm = gm1 = gm2 = 50uA / 25mV = 2mA/V
Avd = Gm * Ro = 2mA * 1.6Meg = 3200 V/V

If the next stage has an input resistance of 1Meg,
Avd = Gm * (Ro || 1Meg) = 1231 V/V

Gm =  (I / 2)  / V_T = 5mA / V
thus, I = 250 uA
R = (5 - (-5) - Vbe) / I
    = 9.3 / 0.25mA
    = 37.2K

Rid = (1+B) 2 re
re = V_T / (I/2) = 25mV / 125uA
    = 200 ohms
thus, Rid = 151 * 2 * 200 = 60.4K