The data field contains the top of the Merkle tree and that is all that is required to solve the block. You don't know what transactions were included in the block you are given to solve.
So the bitcoin daemon gives you the merkle root? From the above posts it sounded like you could decide for yourself what transactions to include in the calculation of the merkle root. But now it sounds like the daemon decides for you.
The CheckBlock method in the bitcoin source only checks that vector of transactions is non-empty, that the first transaction is the coinbase, and that each subsequent transaction is not the coinbase and is a valid transaction. If there is only one transaction and that transaction is the coinbase then the block is valid. So it seems that no matter what merkle root the daemon gives you, if you change it to be the merkle root of only the coinbase transaction you can still legitimately solve the block.
// First transaction must be coinbase, the rest must not be
if (vtx.empty() || !vtx[0].IsCoinBase())
return error("CheckBlock() : first tx is not coinbase");
for (int i = 1; i < vtx.size(); i++)
if (vtx[i].IsCoinBase())
return error("CheckBlock() : more than one coinbase");
// Check transactions
BOOST_FOREACH(const CTransaction& tx, vtx)
if (!tx.CheckTransaction())
return error("CheckBlock() : CheckTransaction failed");