Clipping of a negative feedback amplifier
output stage is the PNP with 1mA sink, not NPN.
if Ve =-2.5V and we assume Vbe of 0.6 with IE=0 causing saturation of input left NPN such that Vce =~0


then \$V_L* R1/(R1+R2)=Vb\$
for \$V_L=+/-1.5V solve for RL

if -1.5V and 1mA sink with 11K bias to gnd + Ibe and Iload , is this what is asked? what is Iload max? 1.5V/11k=0.136mA thus -1.5v/0.863mA =1.73K min R load