Yes if you know the room temperature and the temperature of the cooling object at two known times ##t_{2}>t_{1}>0## after the initial time ##t_{0}=0##.

I will do this in the general case. Somebody else may like to include a specific example (with specific numbers).

Suppose the steady room temperature is ##T_{room}## the temperature of the cooling object at time ##t_{1}>0## is ##H_{1}## where ##T_{room} < H_{1}## and the temperature of the cooling object at time ##t_{2}>t_{1}>0## is ##H_{2}## where ##T_{room} < H_{2} < H_{1}##.
In this situation can be written as giving the temperature ##T## of the object as a function of time ##t## since the object first started cooling in the following way:
##T=f(t)=T_{room}+C*e^{ -k*t}## where the constants ##C>0## and ##k>0## must be found from the given data (that ##f(t_{1})=H_{1}## and ##f(t_{2})=H_{2}##).

This gives two equations in two unknowns (##C## and ##k## are the unknowns): ##H_{1}=T_{room}+C*e^{ -k*t_{1})## and ##H_{2}=T_{room}+C*e^{ -k*t_{2}}##.

To solve this system of equations you could solve the first equation for ##C## in terms of ##k## to get ##C=(H_{1}-T_{room})*e^{k*t_{1}}## and the substitute this into the second equation to get

##H_{2}=T_{room}+(H_{1}-T_{room}) * e^{k*(t_{1}-t_{2})}##

This equation can then be solved for ##k## by using logarithms:

##e^{k*(t_{1}-t_{2})}=(H_{2}-T_{room})/(H_{1}-T_{room})##

##Rightarrow k=(ln((H_{2}-T_{room})/(H_{1}-T_{room})))/(t_{1}-t_{2})=(ln(H_{2}-T_{room})-ln(H_{1}-T_{room}))/(t_{1}-t_{2})## (the numerator and denominator here would both be negative but the fraction would be positive)

We can then ultimately solve for ##C## by substituting into ##C=(H_{1}-T_{room})*e^{k*t_{1}}## to get:

##C=(H_{1}-T_{room})*e^{((ln(H_{2}-T_{room})-ln(H_{1}-T_{room}))/(t_{1}-t_{2}))*t_{1}}##

Finally you can solve for the initial temperature ##H_{0}## as

##H_{0}=f(0)=T_{room}+Ce^{0}=T_{room}+C##

##=T_{room}+(H_{1}-T_{room})*e^{((ln(H_{2}-T_{room})-ln(H_{1}-T_{room}))/(t_{1}-t_{2}))*t_{1}}##

All this would be easier of course if we had specific numbers. But this shows that it is possible to do it in general.