Ошибка при генерации печатной формы

При генеции печатной формы через скрипт в своей системе

 

​var reportService = new Terrasoft.Configuration.ReportService.ReportService();
Terrasoft.Configuration.ReportService.ReportData report = reportService.GenerateMSWordReport(
    (PrintableId.ToString()), ObjectId.ToString(), ConvertToPdf);
var entity = UserConnection.EntitySchemaManager.GetInstanceByName("ActivityFile");
var fileEntity = entity.CreateEntity(UserConnection);
fileEntity.SetDefColumnValues();
fileEntity.SetColumnValue("ActivityId", AddActivityId);
fileEntity.SetColumnValue("TypeId", AttachmentType);
fileEntity.SetColumnValue("Name", "Счет.docx");
fileEntity.SetColumnValue("Data", report.Data);
fileEntity.Save();
return true;
 
​
 
​
 
​

получаю ошибку

System.IO.FileFormatException: File contains corrupted data.
   at MS.Internal.IO.Zip.ZipIOEndOfCentralDirectoryBlock.FindPosition(Stream archiveStream)
   at MS.Internal.IO.Zip.ZipIOEndOfCentralDirectoryBlock.SeekableLoad(ZipIOBlockManager blockManager)
   at MS.Internal.IO.Zip.ZipArchive..ctor(Stream archiveStream, FileMode mode, FileAccess access, Boolean streaming, Boolean ownStream)
   at MS.Internal.IO.Zip.ZipArchive.OpenOnStream(Stream stream, FileMode mode, FileAccess access, Boolean streaming)
   at System.IO.Packaging.ZipPackage..ctor(Stream s, FileMode mode, FileAccess access, Boolean streaming)
   at System.IO.Packaging.Package.Open(Stream stream, FileMode packageMode, FileAccess packageAccess, Boolean streaming)
   at DocumentFormat.OpenXml.Packaging.OpenXmlPackage.OpenCore(Stream stream, Boolean readWriteMode)
   at DocumentFormat.OpenXml.Packaging.WordprocessingDocument.Open(Stream stream, Boolean isEditable, OpenSettings openSettings)
   at Terrasoft.Configuration.ReportService.ReportService.GenerateMSWordReport(String urlTemplateId, String urlRecordUId, Boolean convertInPDF)
   at Terrasoft.Core.Process.elcUsrlabGenerateInvoiceAndSend21Custom31Custom3.ScriptTask1Execute(ProcessExecutingContext context)
   at Terrasoft.Core.Process.ProcessFlowElement.Execute(ProcessExecutingContext context)

Это же решение на демо версии работает корректно.

Подскажите, пожалуйста, куда копать?