The output of the given circuit in figure.

Explanation

Here `beta=100=(I_(C ))/(I_(B)) rArr I_(C )= beta I_(B) " " …(1)`
but `alpha=(beta)/(1+beta)`
`therefore (I_(C ))/(I_(E ))=(100)/(101) rArr I_(C ) ~~I_(E )`
`therefore I_(C )=I_(E ) beta I_(B)" "…(2)`
In output circuit,
`V_(C C)-I_(C )R_(C )-V_(CE)-I_(E )R_(E )=0`
`therefore 12-(beta I_(B))R_(C )-3-(beta I_(B)) (1xx10^(3))=0`
`therefore 9=(beta I_(B)) (R_(C )+1000)`
`therefore 9=(100I_(B))(R_(C )+1000)" "...(3)`
Also,
`V_(B)-V_(BE)-I_(E )R_(E )+(20xx10^(3))I =V_(B)`
`therefore (20xx10^(3))I =V_(BE)+I_(E )R_(E )`
`therefore 20000I =0.5+ beta I_(B)xx(1xx10^(3))`
`therefore 20000 I = 0.5+1000 beta I_(B)`
`therefore 20000 I = 0.5+100000 I_(B) " " ...(4) (because beta = 100)`
For the closed path containing `R_(B)` and R,
`V_(C C)-(I+I_(B))R_(B)-IR=0`
`therefore 12=(I+I_(B)) (100xx10^(3))+(I) (20xx10^(3))`
`therefore 12=120000 I+100000I_(B)" " ...(5)`
Subtracting equation (4) from equation (5),
`12-20000I =120000I-0.5`
`therefore 12.5=140000I`
`therefore I=8.93xx10^(-5)A " "...(6)`
From equation (4) and (6),
`(20000)(8.93xx10^(-5))=0.5+(10^(5)I_(B))`
`therefore 1.786=0.5+(10^(5)I_(B))`
`therefore 10^(5)I_(B)=1.286`
`therefore I_(B)=1.286xx10^(-5)A " "...(7)`
From equation (3) and (7),
`9=(100xx1.286xx10^(-5))(R_(C )+1000)`
`therefore R_(C )+1000=(9)/(1.286xx10^(-3))=(9000)/(1.286)`
`therefore R_(C )+1000=6998.4`
`therefore R_(C )=5998.4Omega`
`therefore ~~6000Omega`
`therefore R_(C )~~6kOmega`