First, you need to work out the amount of HNO3, using n=cv. Make sure to convert the cm3 into dm3 first (to get from cm3 to dm3, divide by 1000).
Once you know this, look at the molar ratio in the reaction (how much of the chemical is on either side). For every 2mol of HNO3, you produce 1mol of Pb(NO3)2. This means that the amont of lead(II) nitrate will be half the amount of nitric acid.
You'll also need the molar mass (Mr) of lead(II) nitrate. Work this out from the periodic table (you should know how to do this).
You can then work out the mass using m = n x Mr. This is called the maximum mass because the mass produced will probably be less than this in an experiment, for example because not all of the reactants would fully react.