Let us 

The gain of a Common Source JFET amplifier is given as

$A_V=-g_mR_D$

Hence the transconductance can be calculated as

$g_m=\dfrac{-A_V}{R_D}=-0.0833$

The drain current is given by the equation

$I_D=\dfrac{V_G}{R_D}$

Voltage between gate and ground is given by the equation

$V_G=R_LI_D=360V$

Applying KVL  at the Drain  Source Junction we have 

$V_{DS}=V_{DD}-I_D(R_D+R_S)$

$V_{DS}=20-(3\times 10^{-3})(R_D+120\times10^3)$

$V_{DS}=20-(3\times10^{-3}R_D-360)$

 Hence the drain resistance can be given by

$R_D=\dfrac{(364.5-20)\times10^3}{3}=114.83K\Omega$

The input impedance is given by the fromula

$R_{in}=R_D$

Hence by the above calculated value we have the input impedance as

$R_{in}=114.83K\Omega$

The Output impedance of Unbypassed Resistor is givenbythe formula

$R_{out}=R_{L}$

Since the Value of Load Resistor is $120K\Omega$

Hence the Output impedance is given bythe formula

$R_{out}=120K\Omega$

The Output Voltage can be calculated from the gain as

$A_v=\dfrac{V_{out}}{V_{in}}$ 

Since Av is 10 we have

$V_{out}=A_vV_{in}$

The Output Voltage is hence calculated as

$V_{out}=10 \times 20V_{pp}=10\times20 \times4.5=90V$